Page 1 Exercise 4.5 page: 4.23 1. Without drawing a diagram, find (i) 10 th square number (ii) 6 th triangular number Solution: (i) 10 th square number The square number can be remembered using the following rule Nth square number = n × n So the 10 th square number = 10 × 10 = 100 (ii) 6 th triangular number The triangular number can be remembered using the following rule Nth triangular number = n × (n + 1)/ 2 So the 6 th triangular number = 6 × (6 + 1)/ 2 = 21 2. (i) Can a rectangular number also be a square number? (ii) Can a triangular number also be a square number? Solution: (i) Yes. A rectangular number can also be a square number. Example – 16 is a rectangular number which can also be a square number. (ii) Yes. A triangular number can also be a square number. Example – 1 is a triangular number which can also be a square number. 3. Write the first four products of two numbers with difference 4 starting from in the following order: 1, 2, 3, 4, 5, 6, ……… Identify the pattern in the products and write the next three products. Solution: We know that 1 × 5 = 5 2 × 6 = 12 3 × 7 = 21 4 × 8 = 32 So the first four products of two numbers with difference 4 5 – 1 = 4 6 – 2 = 4 7 – 3 = 4 8 – 4 = 4 4. Observe the pattern in the following and fill in the blanks: 9 × 9 + 7 = 88 Page 2 Exercise 4.5 page: 4.23 1. Without drawing a diagram, find (i) 10 th square number (ii) 6 th triangular number Solution: (i) 10 th square number The square number can be remembered using the following rule Nth square number = n × n So the 10 th square number = 10 × 10 = 100 (ii) 6 th triangular number The triangular number can be remembered using the following rule Nth triangular number = n × (n + 1)/ 2 So the 6 th triangular number = 6 × (6 + 1)/ 2 = 21 2. (i) Can a rectangular number also be a square number? (ii) Can a triangular number also be a square number? Solution: (i) Yes. A rectangular number can also be a square number. Example – 16 is a rectangular number which can also be a square number. (ii) Yes. A triangular number can also be a square number. Example – 1 is a triangular number which can also be a square number. 3. Write the first four products of two numbers with difference 4 starting from in the following order: 1, 2, 3, 4, 5, 6, ……… Identify the pattern in the products and write the next three products. Solution: We know that 1 × 5 = 5 2 × 6 = 12 3 × 7 = 21 4 × 8 = 32 So the first four products of two numbers with difference 4 5 – 1 = 4 6 – 2 = 4 7 – 3 = 4 8 – 4 = 4 4. Observe the pattern in the following and fill in the blanks: 9 × 9 + 7 = 88 98 × 9 + 6 = 888 987 × 9 + 5 = 8888 9876 × 9 + 4 = ……… 98765 × 9 + 3 = ……… 987654 × 9 + 2 = ………. 9876543 × 9 + 1 = ………. Solution: 9 × 9 + 7 = 88 98 × 9 + 6 = 888 987 × 9 + 5 = 8888 9876 × 9 + 4 = 88888 98765 × 9 + 3 = 888888 987654 × 9 + 2 = 8888888 9876543 × 9 + 1 = 88888888 5. Observe the following pattern and extend it to three more steps: 6 × 2 – 5 = 7 7 × 3 – 12 = 9 8 × 4 – 21 = 11 9 × 5 – 32 = 13 ….. × ……. - ….. = ……. ….. × ……. - ….. = ……. ….. × ……. - ….. = ……. Solution: 6 × 2 – 5 = 7 7 × 3 – 12 = 9 8 × 4 – 21 = 11 9 × 5 – 32 = 13 10 × 6 - 45 = 15 11 × 7 - 60 = 17 12 × 8 - 77 = 19 6. Study the following pattern: 1 + 3 = 2 × 2 1 + 3 + 5 = 3 × 3 1 + 3 + 5 + 7 = 4 × 4 Page 3 Exercise 4.5 page: 4.23 1. Without drawing a diagram, find (i) 10 th square number (ii) 6 th triangular number Solution: (i) 10 th square number The square number can be remembered using the following rule Nth square number = n × n So the 10 th square number = 10 × 10 = 100 (ii) 6 th triangular number The triangular number can be remembered using the following rule Nth triangular number = n × (n + 1)/ 2 So the 6 th triangular number = 6 × (6 + 1)/ 2 = 21 2. (i) Can a rectangular number also be a square number? (ii) Can a triangular number also be a square number? Solution: (i) Yes. A rectangular number can also be a square number. Example – 16 is a rectangular number which can also be a square number. (ii) Yes. A triangular number can also be a square number. Example – 1 is a triangular number which can also be a square number. 3. Write the first four products of two numbers with difference 4 starting from in the following order: 1, 2, 3, 4, 5, 6, ……… Identify the pattern in the products and write the next three products. Solution: We know that 1 × 5 = 5 2 × 6 = 12 3 × 7 = 21 4 × 8 = 32 So the first four products of two numbers with difference 4 5 – 1 = 4 6 – 2 = 4 7 – 3 = 4 8 – 4 = 4 4. Observe the pattern in the following and fill in the blanks: 9 × 9 + 7 = 88 98 × 9 + 6 = 888 987 × 9 + 5 = 8888 9876 × 9 + 4 = ……… 98765 × 9 + 3 = ……… 987654 × 9 + 2 = ………. 9876543 × 9 + 1 = ………. Solution: 9 × 9 + 7 = 88 98 × 9 + 6 = 888 987 × 9 + 5 = 8888 9876 × 9 + 4 = 88888 98765 × 9 + 3 = 888888 987654 × 9 + 2 = 8888888 9876543 × 9 + 1 = 88888888 5. Observe the following pattern and extend it to three more steps: 6 × 2 – 5 = 7 7 × 3 – 12 = 9 8 × 4 – 21 = 11 9 × 5 – 32 = 13 ….. × ……. - ….. = ……. ….. × ……. - ….. = ……. ….. × ……. - ….. = ……. Solution: 6 × 2 – 5 = 7 7 × 3 – 12 = 9 8 × 4 – 21 = 11 9 × 5 – 32 = 13 10 × 6 - 45 = 15 11 × 7 - 60 = 17 12 × 8 - 77 = 19 6. Study the following pattern: 1 + 3 = 2 × 2 1 + 3 + 5 = 3 × 3 1 + 3 + 5 + 7 = 4 × 4 1 + 3 + 5 + 7 + 9 = 5 × 5 By observing the above pattern, find (i) 1 + 3 + 5 + 7 + 9 + 11 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 (iii) 21 + 23 + 25 + ….. + 51 Solution: (i) 1 + 3 + 5 + 7 + 9 + 11 By using the pattern 1 + 3 + 5 + 7 + 9 + 11 = 6 × 6 = 36 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 By using the pattern 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 × 8 = 64 (iii) 21 + 23 + 25 + ….. + 51 We know that 21 + 23 + 25 + ….. + 51 can be written as (1 + 3 + 5 + 7 + ….. + 49 + 51) – (1 + 3 + 5 + …… + 17 + 19) By using the pattern (1 + 3 + 5 + 7 + ….. + 49 + 51) = 26 × 26 = 676 (1 + 3 + 5 + …… + 17 + 19) = 10 × 10 = 100 So we get 21 + 23 + 25 + ….. + 51 = 676 – 100 = 576 7. Study the following pattern: 1 × 1 + 2 × 2 = (2 × 3 × 5)/ 6 1 × 1 + 2 × 2 + 3 × 3 = (3 × 4 × 7) / 6 1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 = (4 × 5 × 9)/ 6 By observing the above pattern, write next two steps. Solution: By using the pattern 1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 On further calculation = (5 × 6 × 11)/6 So we get = 55 By using the pattern 1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 On further calculation = (6 × 7 × 13)/6 So we get = 91 8. Study the following pattern: 1 = (1 × 2)/ 2 1 + 2 = (2 × 3)/ 2 Page 4 Exercise 4.5 page: 4.23 1. Without drawing a diagram, find (i) 10 th square number (ii) 6 th triangular number Solution: (i) 10 th square number The square number can be remembered using the following rule Nth square number = n × n So the 10 th square number = 10 × 10 = 100 (ii) 6 th triangular number The triangular number can be remembered using the following rule Nth triangular number = n × (n + 1)/ 2 So the 6 th triangular number = 6 × (6 + 1)/ 2 = 21 2. (i) Can a rectangular number also be a square number? (ii) Can a triangular number also be a square number? Solution: (i) Yes. A rectangular number can also be a square number. Example – 16 is a rectangular number which can also be a square number. (ii) Yes. A triangular number can also be a square number. Example – 1 is a triangular number which can also be a square number. 3. Write the first four products of two numbers with difference 4 starting from in the following order: 1, 2, 3, 4, 5, 6, ……… Identify the pattern in the products and write the next three products. Solution: We know that 1 × 5 = 5 2 × 6 = 12 3 × 7 = 21 4 × 8 = 32 So the first four products of two numbers with difference 4 5 – 1 = 4 6 – 2 = 4 7 – 3 = 4 8 – 4 = 4 4. Observe the pattern in the following and fill in the blanks: 9 × 9 + 7 = 88 98 × 9 + 6 = 888 987 × 9 + 5 = 8888 9876 × 9 + 4 = ……… 98765 × 9 + 3 = ……… 987654 × 9 + 2 = ………. 9876543 × 9 + 1 = ………. Solution: 9 × 9 + 7 = 88 98 × 9 + 6 = 888 987 × 9 + 5 = 8888 9876 × 9 + 4 = 88888 98765 × 9 + 3 = 888888 987654 × 9 + 2 = 8888888 9876543 × 9 + 1 = 88888888 5. Observe the following pattern and extend it to three more steps: 6 × 2 – 5 = 7 7 × 3 – 12 = 9 8 × 4 – 21 = 11 9 × 5 – 32 = 13 ….. × ……. - ….. = ……. ….. × ……. - ….. = ……. ….. × ……. - ….. = ……. Solution: 6 × 2 – 5 = 7 7 × 3 – 12 = 9 8 × 4 – 21 = 11 9 × 5 – 32 = 13 10 × 6 - 45 = 15 11 × 7 - 60 = 17 12 × 8 - 77 = 19 6. Study the following pattern: 1 + 3 = 2 × 2 1 + 3 + 5 = 3 × 3 1 + 3 + 5 + 7 = 4 × 4 1 + 3 + 5 + 7 + 9 = 5 × 5 By observing the above pattern, find (i) 1 + 3 + 5 + 7 + 9 + 11 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 (iii) 21 + 23 + 25 + ….. + 51 Solution: (i) 1 + 3 + 5 + 7 + 9 + 11 By using the pattern 1 + 3 + 5 + 7 + 9 + 11 = 6 × 6 = 36 (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 By using the pattern 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 8 × 8 = 64 (iii) 21 + 23 + 25 + ….. + 51 We know that 21 + 23 + 25 + ….. + 51 can be written as (1 + 3 + 5 + 7 + ….. + 49 + 51) – (1 + 3 + 5 + …… + 17 + 19) By using the pattern (1 + 3 + 5 + 7 + ….. + 49 + 51) = 26 × 26 = 676 (1 + 3 + 5 + …… + 17 + 19) = 10 × 10 = 100 So we get 21 + 23 + 25 + ….. + 51 = 676 – 100 = 576 7. Study the following pattern: 1 × 1 + 2 × 2 = (2 × 3 × 5)/ 6 1 × 1 + 2 × 2 + 3 × 3 = (3 × 4 × 7) / 6 1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 = (4 × 5 × 9)/ 6 By observing the above pattern, write next two steps. Solution: By using the pattern 1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 On further calculation = (5 × 6 × 11)/6 So we get = 55 By using the pattern 1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 On further calculation = (6 × 7 × 13)/6 So we get = 91 8. Study the following pattern: 1 = (1 × 2)/ 2 1 + 2 = (2 × 3)/ 2 1 + 2 + 3 = (3 × 4)/ 2 1 + 2 + 3 + 4 = (4 × 5)/ 2 By observing the above pattern, find (i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 (ii) 50 + 51 + 52 + ……. + 100 (iii) 2 + 4 + 6 + 8 + 10 + …….. + 100 Solution: (i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 We get = (10 × 11)/2 On further calculation = 55 (ii) 50 + 51 + 52 + ……. + 100 We can write it as (1 + 2 + 3 + …… + 99 + 100) – (1 + 2 + 3 + 4 + ….. + 47 + 49) So we get (1 + 2 + 3 + …… + 99 + 100) = (100 × 101)/2 (1 + 2 + 3 + 4 + ….. + 47 + 49) = (49 × 50)/2 By substituting the values 50 + 51 + 52 + ……. + 100 = (100 × 101)/2 + (49 × 50)/2 On further calculation = 5050 – 1225 We get = 3825 (iii) 2 + 4 + 6 + 8 + 10 + …….. + 100 We can write it as 2 (1 + 2 + 3 + 4 + …… + 49 + 50) So we get = 2 × (50 × 51)/2 On further calculation = 2 × (1275) We get = 2550Read More

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